4 Answers
4

You have:
$$\int_{e}^{2e}\frac{x}{\log x}dx = e^2\int_{1}^{2}\frac{x}{1+\log x}dx=e^2\int_{0}^{\log 2}\frac{e^{2x}}{1+x}dx=\int_{1}^{1+\log 2}\frac{e^{2x}}{x}dx.$$
Despite the fact that no elementary antiderivative exists, you can exploit the fact that $\frac{e^{2x}}{1+x}$ is a very regular function on $[0,\log 2]$, and integrate termwise its Taylor series in $x=0$. We have:
$$\begin{eqnarray*}\int_{1}^{1+\log2}\frac{e^{2x}}{x}dx &=& \log(1+\log 2)+\int_{1}^{1+\log 2}\frac{e^{2x}-1}{x}dx\\&=&\log(1+\log 2)+\left[\sum_{j=1}^{+\infty}\frac{(2x)^{j}}{j\cdot j!}\right]_{1}^{1+\log 2}\\&=&\log(1+\log 2)+\sum_{j=1}^{+\infty}\frac{2^j((1+\log 2)^j-1)}{j\cdot j!}\\&=&\log(1+\log 2)+\sum_{j=1}^{+\infty}\frac{2^j}{j\cdot j!}\sum_{k=1}^{j}(\log 2)^k\\&=&\log(1+\log 2)+\sum_{k=1}^{+\infty}(\log 2)^k\sum_{j\geq k}\frac{2^j}{j\cdot j!}.\end{eqnarray*}$$
Another efficient technique to calculate such an integral numerically is to exploit the continued fraction representation for the exponential integral.

This is not an elementary integral. Let $x=e^{-z}$ and observe that the integral be written in terms of the exponential integral, defined as $$\text{Ei}(t) = -\int_{-t}^\infty \frac{e^{-z}}{z}\,dz$$ which as noted on the Wikipedia page is not an elementary function.

The change of variables $x=e^y$ gives the integral $\int \frac{e^{2y}}ydy$, which is known not to be computable in elementary functions. Thus, the answer can be only expressed via the exponential integral,
$\int...=Ei(2\ln2\epsilon)-Ei(2\ln\epsilon)$.